I'm working through some algebra qualifying exam questions, and came across the following problem which has totally stumped me.
Let $n \in \mathbb{N}$.
a) Explicitly construct a field extension $K/\mathbb{Q}$ such that $\text{Gal}(K/\mathbb{Q}) \cong \mathcal{C}_n$.
b) Find an explicit algebraic number $\alpha$ such that $\text{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q}) \cong \mathcal{C}_3$.
So part b seems to follow immediately if I can figure out part a. The problem I'm having here is that the only results from the textbook I'm using on cyclic extensions require you to have an $n$th primitive root of unity $\zeta_n$ already sitting inside of the base field.
Hence, the only idea I have is somehow make a sufficiently large cyclotomic extension $F_{/\mathbb{Q}}$ with intermediate field $K$ such that $\text{Gal}(F_{/K}) \lhd \text{Gal}(F_{/\mathbb{Q}})$ such that $\text{Gal}(F_{/\mathbb{Q}}) / \text{Gal}(F_{/K}) \cong \mathcal{C}_n$.
However, this approach doesn't seem to help me for b. I know how to easily show existence of such an $\alpha$. Simply take an separable cubic over $\mathbb{Q}$ with discriminant being a square in $\mathbb{Q}$, and then apply the primitive element theorem to the splitting field of said cubic. However, I have no idea how to find $\alpha$ explicitly.